1. The document discusses the second law of thermodynamics and concepts related to entropy, including spontaneous and non-spontaneous processes, the Carnot cycle, entropy changes in reversible and irreversible processes, statements of the second law, and free energy functions. 2. It introduces the Carnot cycle as a model for converting heat into work using an ideal gas as a working substance through four steps of isothermal and adiabatic changes. 3. Entropy is defined in relation to reversible processes as the ratio of heat absorbed to temperature (q/T). The second law is explained through entropy changes and the principle that the total entropy change is zero for reversible processes but increases for irreversible processes.

1. 1
Thermodynamics
Part - II
Mr.M.RAGU
Assistant Professor of Chemistry
Vivekananda College
Tiruvedakam West – 625234
Madurai -Tamilnadu

2. THE SECOND LAW OF
THERMODYNAMICS
Spontaneous and Non-Spontaneous processes
The physical or chemical changes which proceed by themselves
without the intervention of any external agency are known as
“spontaneous processes.” All natural processes are spontaneous.
They tend to attain an equilibrium. All spontaneous processes
proceed only in one direction and are thermodynamically
irreversible.
Examples
i) Water flows by itself from a higher level to a lower level
ii) Heat flows from a hotter to a colder body till they attain
thermal equilibrium
iii) Dffusion of a gas or solute takes place from higher to lower
concentration
iv) The expansion of a gas into an evacuated space is spontaneous

3. By the use of an external agency it is possible to
bring about the reversal of a spontaneous process. For
example, it is possible to make water flow from a lower
to a higher level with the aid of an external agency (water
pump). These process which proceed in both the
direction are called “non-spontaneous” or “reversible
processes.”
All natural processes are spontaneous and
thermodynamically feasible . E.g. The flow of heat from a
hot body to a cold body takes place spontaneously. But
the reverse process of flow of heat from a cold body to a
hot body does not take place spontaneously. It is said to
be thermodynamically not feasible.

4. The need for a second law in thermodynamics
i) The first law of thermodynamics tells us that energy can be changed from
one form to another. It is possible to convert heat into other forms of
energy. This law does not say anything about the source of heat and the
direction of its flow. It cannot predict whether heat can flow
spontaneously from a cold body to a hot body . All we can say from the
first law is that the heat lost by the hot body is the same as the heat
gained by the cold body.
ii) Although the first law permits the inter-conversion of energy, it says
nothing about the extent of convertibility of one form of energy into
another. For example, mechanical energy can be converted into an
equivalent amount of electrical energy. When work is done on a gas , an
equivalent amount of heat would be produced. But heat cannot be
completely converted into work.
iii) All natural processes which occur spontaneously tend to acquire a state of
minimum energy by the evolution of heat. Thus, the function H
introduced by the first law predicts that exothermic reactions with a
decrease in ΔH(ie – ΔH) are spontaneous. However, many endothermic
reactions with increse in ΔH(ie+ ΔH) do take place spontaneously. Hence,
ΔH is not sufficient to predict the direction and feasibility of a process.
Thus, it is obvious from the above discussion that the first law is
inadequate to assess how far and in which direction a reaction would
proceed and an additional law is required.

5. The Carnot Cycle
(Conversion of heat into work)
Any device that can convert heat into work is called a “heat engine”.
S.Carnot devised a heat engine using an ideal gas as the working
substance (system). The gas is subjected to a cyclic change which involves
four steps:
1. Isothermal reversible expansion
2. Adiabatic reversible expansion
3. Isothermal reversible compression
4. Adiabatic reversible compression

6. Step 1 (AB) : The gas absorbs q1 calories of heat from the source and gets
expanded “isothermally” and “reversibly” from V1 to V2 at temperature T1.
The work done by the gas is
w1 = RT1 ln V2/V1
Step 2 (BC): The gas is expanded “adiabatically” and “reversibly” from V2 to
V1 so that the temperature falls from T1 to T2. During the process, q=0 and
the work done by the gas is
w2 = -CV ( T2-T1)
Step 3 (CD): The gas is compressed “isothermally” and “reversibly” from V3
to V4 at temperature T2 with the liberation of q2 calories of heat to the sink.
The work done by the gas is
w3 = RT2 ln V4/V3
Step 4 (DA): The gas is compressed “adiabatically” and “reversibly” so that
it returns to the initial volume V1 from V4 that the temperature increases
from T2 to T1. During the process, q=0 and the work done by the gas is
w4 = -CV ( T1-T2)
= CV ( T2-T1)
The total work done by the engine is
W = w1 + w2 + w3 + w4
= RT1 ln V2/V1 - CV ( T2-T1) + RT2 ln V4/V3 + CV ( T2-T1)
= RT1 ln V2/V1 + RT2 ln V4/V3

7. Since V2/V1 = V3/V4
Hence w = R(T1 - T2)ln V2/V1
The net heat absorbed(q) by the ideal gas in the whole cycle is given by
q = q1- q2 = RT1 ln V2/V1 + RT2 ln V4/V3
RT1 ln V2/V1 - RT2 ln V3/V4
Since in adiabatic expansion of an ideal gas
CV ln(T1/T2) = R ln V3/V2 (For Step 2)
CV ln(T1/T2) = R ln V4/V1 (For Step 4)
or V3/V2 = V4/V1
or V2/V1 = V3/V4
Hence the net heat absorbed
q = q1- q2 = R(T1 - T2)ln V2/V1
This is equal to the net work done by the engine ie q = w . Thus, the essential
condition for a cyclic process that net work done is equal to net heat
absorbed is fully satisfied.
The relationship between w, the net work done by the system and q2, the
quantity of heat adsorbed at the higher temperature T1. in case of Carnot
cycle can be obtained from the following two equations:
w = R(T1 - T2)ln V2/V1
and q1 = RT1 ln V2/V1
Hence, w = q2 (T1 – T2)/T1
Since (T1 – T2)/T1 < 1, it follows that w< q2. This means that only a part of the
heat absorbed by the system at the higher temperature T1 is transformed into
work. The rest of the heat (q2) is given out by the system to the surroundings
when it is at lower temperature T2.

8. By definition, the efficiency of the engine is
η = Total work done by the gas/ Heat absorbed from the source
= w/q1
= R(T1 - T2)ln V2/V1 /RT1 ln V2/V1
=(T1 - T2)/T1
From the expression for the efficiency of a Carnot engine , it is seen
that
i) η will be unity when T2=0 ie when the temperature of the system is
absolute zero. This is not attainable. Hence, T1 is always greater than
T2 and η will be less than unity. This means that the heat absorbed
from the source is not completely converted into work.
ii) The efficiency of a heat engine depends only on the temperatures
of the source and the system is independent of the nature of the
working substance. This is called “Carnot-Clasius Theorem”.
iii) All reversible engines working between the same temperatures of
the source and the system will have the same efficiency . This is
known as “Carnot Theorem”.

9. Entropy
⚫ The efficiency of a Carnot engine is given by
⚫η =(T1 - T2)/T1 =(q1 – q2)/q1
⚫or 1- (q2/q1) = 1- (T2/T1)
⚫or (q2/q1) = (T2/T1)
⚫ ჻ (q2/T2) = (q1/T1)
⚫In general,
⚫q/T= constant for a reversible cyclic process. If q2 is the
heat absorbed and q1 is the heat given out, q2 is +ve and q1
is –ve. Now
⚫ (q2/T2) =- (q1/T1)
⚫or (q2/T2) + (q1/T1) = 0

10. A reversible cycle may be regarded as being made up of a number
of Carnot cycles. In such a case, the sum of q/T terms is equal to
zero.
Σq/T = 0
When the changes are infinitesimal,
Σδq/T = 0
In the limit, the summation can be replaced by a cyclic integral
Σδq/T = 0
ʃδq/T = 0
Since the cyclic integaral is zero, then the əq/T has the qualities of a
thermodynmic state function. Clausius called this function “entropy”.
The entropy change accompanying a reversible isothermal change is
given as
dS = δq/T
Integrating the above equation between limits
1ʃ2 = 1ʃ2 δq/T
or S2-S1 = ΔS = q/T
Thus, entropy change of a reversible system may be defined as “the
ratio of the heat absorbed and the temperature” in the absolute scale.
Units of Entropy
i) Cal deg-1(e.u)

11. Entropy change for reversible and irreversible processes
⚫For Reversible Process
⚫ If a system undergoes a
reversible change through a series
of Carnot cycles from states A to B,
then the entropy change is given by
⚫ Δssys = qrev/T
⚫ Δssur = -qrev/T
⚫ or Δssys + Δssur = qrev/T -qrev/T=0
⚫ Thus, in a reversible process, the
total entropy change of the system
and the surroundings will be zero.
⚫For Irreversible Process
⚫ Consider a process in which the
absorption of heat by the system
takes place under reversible
condition while the remaining part
of the Carnot cycle is irreversible.
Then the entropy is
⚫ Δssys = qrev/T
⚫ Δssur = -qirr/T
⚫ or Δssys + Δssur = qrev/T -qirr/T
We know that, Wrev > Wirr
჻ qrev > qirr
qrev/T – qirr/T > 0
or Δssys + Δssur > 0
or Δssys ≠ Δssur
Thus, the entropy changes of the system and the surroundings are not the
same. This is the “Clausius inequality principle”.

12. The Second law of Thermodynamics
⚫ The second law of thermodynamics has been stated in
various forms:
⚫1. Kelvin Statement : “It is possible to construct an
engine which is able to convert heat completely into work
by a cyclic process without producing changes either in
the system or in the surroundings”.
⚫2. Clausius statement : “ It is impossible to construct an
engine which is able to convey heat by a cyclic process
from a cold reservoir to a hot reservoir without the aid of
an external agency”.
⚫3. Entropy statement : “The entropy of the universe
remains constant in a reversible process but it tends to
increase in a spontaneous process”.

13. Entropy and Probability
⚫ A system tends to change spontaneously from an ordered state to a
disordered state since disordered states have a much higher probability than
ordered states. This means that all spontaneous processes are accompanied by
an increase of entropy and randomness (disorder). Therfore, entropy may be
regarded as a function of probability of the thermodynamic state and a
measure of disorder. A quantitative relationship between entropy S and
probability w is given as
⚫ S = - k ln w
⚫ where k = Boltzman constant. w being a fraction. Ln w will be negative
and S can have positive value (for spontaneous processes) only if the
expression has a negative sign.
⚫ In this connection it may be noted that gaseous state is more random
than liquid state and liquid state is more random than solid state. Thus,
melting of ice and evaporation of water proceeds spontaneously in the
direction of greater randomness and hence entropy.
⚫ Similarly, expansion of a gas, dissolution of a salt in water and
demagnetisation of iron are associated with an increase of entropy due to
increased randomness.

14. Entropy as a criterion for spontaneity of a reaction
(Physical significance of Entropy)
⚫In all natural or spontaneous processes, the system tends
to proceed in the direction of increased randomness. Since
entropy is a measure of randomness, the + ve sign of ∆S is
a criterion for spontaneity or feasibility of a reaction.
⚫ ∆S > 0
⚫or ∆S = +ve
⚫For a reversible process,
⚫ ∆S = 0
⚫ For example, if we take some water in a beaker, it is seen
to evaporate spontaneously at room temperature. It is due
to increased randomness and entropy (∆S is +ve).

15. Free Energy Functions
⚫ Entropy change is applied as a criterion for
spontaneity of a process. However, it is not a
convenient criterion because one has to compute ∆S
for the system and its surroundings. Hence, it is
desirable to introduce new thermodynamic functions
which concern only with the system as criteria for
spontaneity or feasibility of a process. Two such
functions are derived from the concepts of Energy
(Ilaw) and Entropy(II Law) of the system.
⚫They are:
⚫ 1.Helmholtz free energy, A
⚫2. Gibbs free energy, G

16. Helmholtz Free Energy
⚫ The Helmholtz free energy is defined as
⚫ A = E-TS
⚫ For a small change
⚫ ∆A = ∆E - T ∆S
⚫ By definition, ∆S = q/T
⚫ or T ∆S = q
⚫ Sustituting the value of T∆S
⚫ ∆A = ∆E – q
⚫ According to First law
⚫ ∆E = q-w
⚫ or ∆E – q = -w
⚫ Sustituting this value
⚫ ∆A = -w
⚫ For a reversible process
⚫ ∆A = -wmax
⚫ Thus, the decrease in function A gives the maximum work that can be obtained
during an isothermal and reversible transformation of a system. Wmax represents
the total work including all kinds of work such as work of expansion, electrical
work and gravitational work etc. For this reason the Helmholtz free energy , A is
usually referred to as work function.

17. Gibbs Free Energy
⚫ The Gibbs free energy is defined as
⚫ G = H-TS
⚫ For a small change
⚫ ∆G = ∆H-T ∆S
⚫ But ∆H = ∆E + P∆V
⚫ ჻ ∆G = ∆E - T∆S + P∆V
⚫ By definition,
⚫ ∆A = ∆E - T∆S
⚫ Substituting the value of ∆A
⚫ ∆G = ∆A + P∆V
⚫ For a reversible isothermal change
⚫ ∆A = -wmax
⚫ ჻ ∆G = -wmax + P∆V
⚫ or -∆G = wmax - P∆V
⚫ But P∆V= work function
⚫ = wexpansion
⚫ ჻ - ∆G = wmax - wexpansion
⚫ Thus, the decrease in function G gives the net work other than the work of
expansion that can be obtained from the isothermal reversible
transformation system . The function is called free energy.

18. Relationship between A and G
⚫∆G = ∆A + P ∆V
⚫Distinction between Helmhlotz free energy and
Gibbs free energy
Helmhlotz free
energy
Gibbs free energy
1 It gives the maximum
total work obtainable
from a system.
It gives the net useful
work obtainable from a
system.
2 It includes all kinds of
work.
It includes work other
than mechanical work
3 It originates from the
internal energy of the
system A = E- TS
It originates from the
enthalpy of the system
G = H - TS

19. Physical significance of Free energy
⚫ i) If ∆G is –ve, the reaction is spontaneous and
thermodynamically feasible.
⚫ii) If ∆G is +ve, the reaction does not occur in the
direction indicated by the equation. That is, it is
not thermodynamically possible. However, it shall
occur in the reverse direction.
⚫iii) If ∆G is zero, the system is in a state of
equilibrium. There is no net reaction in either
direction. Such reactions are said to be
thermodynamically not feasible but possible.

20. Free energy change as the true thermodynamic
criterion for spontaneity of a process
i) ∆H and spontaneity
It has been found that most of the reactions which occur
spontaneously at room temperature are exothermic. It is
because of the fact that the reacting system tends to acquire a
state of lower energy by evolution of heat.
∆H = Hproducts – Hreactants
If Hproducts < Hreactants
∆H = -ve (exothermic reaction)
Thus, the -ve ∆H maybe taken as a criterion for spontaneity
or feasibility of a reaction. But it is not valid in all cases. There
are a number of endothermic reactions (∆H is +ve) which are
known to proceed spontaneously. For example, if a small
amount of Ammonium Chloride is added to water, it dissolves
readily. The process is endothermic and yet spontaneous .
Therefore, the sign of ∆H alone does not determine this
spontaneity or feasibility of a process or reaction.

21. 2. ∆S and spontaniety
⚫ All natural or spontaneous processes proceed in the
direction of increased randomness or disorder. For such a
process, the entropy change increases and ∆S is given a
+ve sign. Thus, the +ve sign of ∆S is a criterion for
spontaneity or feasibility of a reaction. But this is not
always true. For example, the combustion of carbon
monoxide is a spontaneous process although it involves a
decrease in entropy (∆S is –ve).
⚫ 2CO(g) + O2(g) → 2CO2(g)
∆S = Sproducts – Sreactants
Sproducts < Sreactants
∆S = -ve
Thus, the sign of ∆S can not be the “sole”criterion for
determinining the spontaneity or feasibility of a reaction.

22. 3. ∆G and spontaneity
⚫ The spontaneity or feasibility of a reaction is
controlled neither by the enthalpy change (energy
factor) alone nor by the entropy change (randomness
factor) alone. The free energy change which is the
resultant of the two factors is the true thermodynamic
criterion for judging the spontaneity or feasibility of a
reaction.
⚫∆G = ∆H - T ∆S
⚫ The sign of ∆G depends upon the signs and the
numerical values of ∆H and T∆S.
⚫ If ∆G is –ve, the reaction is feasible.
⚫If ∆G is +ve, the reaction is not feasible.
⚫If ∆G is zero, the reaction is reversible.

23. Standard free energy change
⚫ The standard free energy of formation, ∆Gf
o is
defined as the “free energy change that results when 1
mole of a compound is formed from the constituent
elements in the standard state of 1 atm pressure and
25oC”. From a knowledge of the ∆Gf
o values of the
products and reactants, the standard free energy
change of a reaction can be calculated.
⚫ ∆Gf
o = Σ∆Gf
o
(products) - ∆Gf
o
(reactants)
⚫ The ∆Go values are useful in predicting the feasibility
of a reaction.

24. Partial molar properties
⚫ For a closed system, the mass and composition remain unchanged.
In such a case, any thermodynamic property X, depends on any two of
the three variables P, V and T. However, in an open system where
transfer of matter is permitted, the mass and composition vary. In
such cases, the thermodynamic property X, depends not only on P, V
and T but also on the quantity of the various components present in
the system. This may be represented as
⚫ X = f(T, P, n1, n2……….ni)
⚫ where n1, n2……….ni represent the number of moles of components 1,
2 etc. For a small change in T and P of the system as well as in the
amounts of the components, the change in property X is given by
⚫ dX = (Əx/ƏT)P,n1,n2….dT + (Əx/ƏP)T,n1,n2….dP +
⚫ (Əx/Ən1)P,T, n2,n3….dn1 + (Əx/Ən2)P, T, n2,n3….dn2
⚫ In the above equation (Əx/Ən1)P,T, n2,n3….dni is the partial derivative of
X with respect to the amount of component 1 when T, P and all the
other n’s are held constant. This is called the “partial molar quantity”
or “partial molar property “with respect to component 1 of the system.

25. Examples
⚫ (ƏV/Əni)P,T,N = Partial molar volume of system
⚫ (ƏG/Əni)P,T,N = Partial molar free energy of a system
⚫ The partial molar property of a system may be defined as “the
change in any thermodynamic property of the system caused when
one mole of a pure component is added to the system keeping
temperature, pressure and composition of all other components
constant”.
⚫ The partial molar property is represented as X, so that
⚫ (Əx/Ən1)P,T, n2,n3… = X1
⚫ (Əx/Ən2)P,T, n1,n3… = X2
⚫ It is thus possible to write the eqn. as
⚫ dX = (Əx/ƏT)P,n1,n2….dT + (Əx/ƏP)T,n1,n2….dP + X1dn1 + X2dn2
⚫ At constant temperature and pressure the first two terms
vanish. Now,
⚫ (dX)T,P = X1dn1 + X2dn2 …….. + Xidni

26. Chemical Potential
⚫ The partial molar properties of the components in a binary mixture may be
related as
⚫ (dX)T,P = X1dn1 + X2dn2
⚫ If the thermodynamic proprty X is the Gibb’s free energy G, then
⚫ (dG)T,P = G1dn1 + G2dn2
⚫ Where G1 = partial molar free energy of component 1
⚫ G2 = partial molar free energy of component 2
⚫ The partial molar free energy of a component is called its “chemical potential”. It is
denoted by μ. The chemical potential is defined as the “change in free energy of a
system associated when one mole of a pure component is added to the system keeping
temperature, pressure and composition of all other components constant”. Since G = μ,
⚫ (dG)T,P = μ1dn1 + μ2dn2
⚫ On integration, (G)T,P = μ1n1 + μ2dn2
⚫ In general , (G)T,P = Σiμini
⚫ Thus, chemical potential may be regarded as the contribution per mole of each
component of the mixture to the total free energy of the system at constant
temperature and pressure.
⚫ Significance: The spontaneity of a chemical reaction can be predicted from partial
molar free energies of the variuos components. Hence, the name chemical
potential.

27. Gibbs-Duhem Equation
⚫ For a binary system,
⚫ (dG)T,P = μ1dn1 + μ2dn2………………………...(1)
⚫ (G)T,P = μ1n1 + μ2dn2 ………………………...(2)
⚫ On differentiation,
⚫ (dG)T,P =(μ1dn1 + n1dμ1)(μ2dn2 + n2dμ2)
⚫ = (μ1dn1 + μ2dn2)(n1dμ1 + n2dμ2) ………………………...(3)
⚫ Comparing Eqns. (1) and (3)
n1dμ1 + n2dμ2 = 0
In general Σ nidμi = 0
This simple equation which relates the change in chemical potential with
composition is called “Gibbs-Duhem equation”.
n1dμ1 + n2dμ2 = 0
⚫ or n1dμ1 = -n2dμ2
⚫ or dμ1 = -n2/n1dμ2
⚫ Thus, the change in chemical potential of component 1 can be
calculated if that of component 2 is known. This means that the
chemical potential does not vary independently.

28. Applications of Gibb’s-Duhem equation
⚫ 1. The Gibbs-Duhem equation finds extensive use in heterogeneous equilibrium particularly liquid-
vapour system.
⚫ 2. It can also be used to calculate the chemical potential of the solute from that of the solvent in a
binary liquid system. The Gibb’s –Duhem equation is
⚫ n1dμ1 + n2dμ2 = 0
⚫ The subscripts 1 and 2 refer to the solvent and solute respectively.
⚫ dμ2 = -n1/n2dμ1
⚫ Since n1/n2 = x1/x2, we have
⚫ dμ2 = -x1/x2dμ1
⚫ where X1 and X2 are the mole fraction of solvent and solute respectively.
⚫ We know,
⚫ μ = μo + RT ln x
⚫ Differentiating at constant T and P for the solvent
⚫ dμ1 = (RT/x1)d x1
⚫ Substituting the value of dμ1
⚫ dμ2 = -(x1/x2)(RT/x1)d x1
⚫ = -RT .(dx1)/x2
⚫ However, x1+x2 = 1, so that -d x1= d x2
⚫ dμ2= RT .(dx2)/x2
⚫ On integration, we have
⚫ μ2 = RT ln x2 + C
⚫ Where C is the constant of integration. Under the standard condition of x2 = 1, μ= μo. Now, the above eqn.
reduces as μ2
o= C. On sustituting this
⚫ μ2 = μ2
o + RT ln x2

29. The Gibbs-Helmhotz Equation
⚫ The free energy change in terms of P and T is given as
⚫ dG = VdP- SdT
⚫ At constant pressure, dP = 0
⚫ ჻ dG = -SdT
⚫ Suppose G1 and G2 are the free energy of a system at the initial and final states
respectively.
⚫ dG1 = -S1dT
⚫ dG1 = -S1dT
⚫ ჻ dG2 - dG1 = -(S2-S1)dT
⚫ or d(G2- G1)= -(S2-S1)dT
⚫ or d∆G= -∆SdT
⚫ or [ə(∆G)/ƏT]P = -∆S
⚫ We know
⚫ ∆G = ∆H - T∆S
⚫ Sustituting the value of ∆S in the eqn.
⚫ ∆G = ∆H + T [ə(∆G)/ƏT]P
⚫ This is the “Gibbs-Helmholtz equation” in terms of enthalpy and free energy
change. For a reaction at constant volume , a similar relation may be derived
in terms of internal energy and work function as
⚫ ∆S = ∆E+ T [ə(∆A)/ƏT]V

30. Integrated form of Gibbs-Helmholtz equation
⚫ The differential form of Gibbs-Helmholtz equation is
⚫ ∆G = ∆H + T [ə(∆G)/ƏT]P
⚫ or - ∆H = T [ə(∆G)/ƏT]P-∆G
⚫ Dividing both sides by T2
⚫ - ∆H/ T2 = {T [ə(∆G)/ƏT]P-∆G}/ T2………………..(1)
⚫ The R.H.S of the above equation may be obtained by differentiating ∆G/T w.r.
to T at constant pressure
⚫ [ə(∆G/T)/ƏT]P = {T. (ə/əT)(∆G) - ∆G. (ə/əT)(T)}/T2
⚫ = {T [ə(∆G)/ƏT]P - ∆G}/T2 ………………..(2)
⚫ Comparing the equations (1) and (2)
⚫ [ə(∆G/T)/ƏT]P = - ∆H/ T2
⚫ or d(∆G/T) = - ∆H.(dT/ T2)
⚫ On integration between the limits
⚫ 1ʃ2d(∆G/T) = - ∆H T1ʃT2 dT/ T2
⚫ (∆G2/T2) -(∆G1/T1) = - ∆H[-1/ T]T2
T1
⚫ = ∆H[1/ T]T2
T1
⚫ = ∆H[(1/ T2)- [(1/T1)]
⚫ = - ∆H[(1/ T1)- [(1/T2)]
⚫ = - ∆H[(T2 - T1)/( T1T2)]
჻ (∆G2/T2) -(∆G1/T1) = - ∆H[(T2 - T1)/( T1T2)]

31. Applications of Gibbs-Helmholtz equation
⚫ 1. The most important application of Gibbs-Helmholtz equation is to calculate the enthalpy
change of reactions taking place in galvanic cells.
⚫ ∆G = ∆H + T [ə(∆G)/ƏT]P
⚫ But ∆G = -nFE
⚫ ჻ -nFE = ∆H + T [ə(-nFE)/əT]P
⚫ = ∆H -nFT (əE/əT)P
⚫ or ∆H = -nFE - nFT (əE/əT)P
⚫ or ∆H = -nF[E - T (əE/əT)P]
⚫ Where E = E.M.F of the cell
⚫ (əE/əT)P = Temperature co-efficient of E.M.F of the cell
⚫ 2. Gibbs-Helmholtz equation enables evaluation of entropy change of cell reactions
⚫ ∆G = ∆H - T∆S
⚫ or ∆G - ∆H = - T∆S
⚫ But ∆G - ∆H = T [ə(∆G)/ƏT]P
⚫ ჻ - T∆S = T [ə(∆G)/ƏT]P
⚫ or - ∆S = [ə(∆G)/ƏT]P
⚫ But ∆G = -nFE
⚫ ჻ - ∆S = [ə(-nFE)/ƏT]P
⚫ or - ∆S = -nF [əE/ƏT]P
⚫ or ∆S = nF [əE/ƏT]P
⚫ 3. It is used to deduce various equations like Clausius-Clapeyron equation, Van’t Hoff equation in
physical chemistry.
⚫ 4. It is helpful for the development of Nernst Heat Theorem, the precursor to the third law of
thermodynamics.
⚫ 5. The integrated form of Gibbs-Helmholtz equation is very useful to calculate the free energy at
any desired temperature.

32. The Clausius-Clapeyron Equation
⚫ The clapeyron equation for phase transition is
⚫ dP/dT = ∆H/T∆V
⚫ For liquid-vapour equilibrium
⚫ ∆V = Vg-Vl
⚫ Since Vg >Vl
⚫ Vg-Vl = Vg
⚫ ჻ dP/dT = ∆Hv/T∆Vg
⚫ For an ideal gas
⚫ PV = RT
⚫ ჻ Vg = RT/P
⚫ Substituting the value of Vg
⚫ dP/dT = (∆HV/T).(P/RT)
⚫ or (1/P)(dP/dT) = (∆HV/RT2)
⚫ or (dlnP/dT) = (∆HV/RT2)
⚫ This expression is known as the differential form of “Clausius-
Clapeyron equation”.

33. Integrated form of Clausius-Clapeyron
equation
⚫ (dlnP/dT) = (∆HV/RT2)
⚫ or dlnP = (∆HV/R).( dT /T2)
⚫On integration between the limits
⚫ P1ʃP2 dlnP = (∆HV/R) T1ʃT2( dT /T2)
⚫ Ln P2/P1 = -(∆HV/R) [-1 /T] T1
T2
⚫ = - ∆HV/R[(1/ T2)- [(1/T1)]
⚫ = ∆HV/R[(1/ T1)- [(1/T2)]
⚫ = ∆HV/R[(T2 - T1)/( T1T2)]
⚫or 2.303 log P2/P1 = ∆HV/R[(T2 - T1)/( T1T2)]
⚫჻ log P2/P1 = (∆HV/2.303R)[(T2 - T1)/( T1T2)]

34. Applications of Clausius – Clapeyron equation
1. Clausius – Clapeyron equation is used to calculate the heat of
vapourisation(∆HV) from the vapour pressures at two
different temperatures. It can also be obtained from the log P
versus 1/T plots. The slope of the straight line obtained is
equal to - ∆HV/2.303R.
2. This equation can be used to study the effect of pressure on
the boiling point of a liquid. The boiling point of a liquid at a
desired pressure can be calculated provided the boiling point
at another pressure is known.
3. It is helpful to study the effect of temperature on the vapour
pressure of a liquid. If the vapour pressure of a liquid at one
temperature is known, its value at another temperature can
be evaluated.
4. It is used to derive expressions for the molal elevation
constant (∆Tb) and molal depression constant (∆Tf) of a
dilute solution.

35. Thermodynamic derivation of Law of mass action
⚫ Consider the general gaseous reaction
⚫ aA + bB → lL + mM
⚫ Let PA, PB, PL and PM be the partial pressures of A, B, L and M.
⚫ Free energy of ‘a’ mole of A} = aGA = aGo
A + aRT ln PA
⚫ Free energy of ‘b’ mole of B} = bGB = bGo
B + bRT ln PB
⚫ Free energy of ‘lc’ mole of L} = lGL = lGo
L + lRT ln PL
⚫ Free energy of ‘m’ mole of M} = mGM = mGo
M + mRT ln PM
⚫ The free energy change of the reaction is
⚫ ∆G = Gproducts – Greactants
⚫ = lGL + mGM - aGA – bGB
⚫ = (lGo
L + lRT ln PL) + (mGo
M + mRT ln PM)
⚫ - (aGo
A + aRT ln PA) - (bGo
B + bRT ln PB)
⚫ On rearranging we have
⚫ ∆G = (lGo
L + mGo
M - aGo
A - bGo
B) + RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ = ∆Go + RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ Where ∆Go is the standard free energy change of the reaction. At equilibrium ∆G = 0 .
⚫ ჻ ∆Go = - RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ Since ∆Go for a particular reaction at a given temperature is a fixed quantity, the logarthimic factor
must have constant value. It is customary to call this conatant as “equilibirium constant”, KP.
⚫ i.e {[(PL)l (PM)m]/[(PA)a (PB)b]} = KP
⚫ Since the partial pressures are proportional to molar concentration.
⚫ {[(L)l (M)m]/[(A)a (B)b]} = KP

36. Van’t Hoff Reaction Isotherm
⚫ Consider the general gaseous reaction
⚫ aA + bB → lL + mM
⚫ Let PA, PB, PL and PM be the partial pressures of A, B, L and M.
⚫ Free energy of ‘a’ mole of A} = aGA = aGo
A + aRT ln PA
⚫ Free energy of ‘b’ mole of B} = bGB = bGo
B + bRT ln PB
⚫ Free energy of ‘lc’ mole of L} = lGL = lGo
L + lRT ln PL
⚫ Free energy of ‘m’ mole of M} = mGM = mGo
M + mRT ln PM
⚫ The free energy change of the reaction is
⚫ ∆G = Gproducts – Greactants
⚫ = lGL + mGM - aGA – bGB
⚫ = (lGo
L + lRT ln PL) + (mGo
M + mRT ln PM)
⚫ - (aGo
A + aRT ln PA) - (bGo
B + bRT ln PB)
⚫ On rearranging we have
⚫ ∆G = (lGo
L + mGo
M - aGo
A - bGo
B) + RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ = ∆Go + RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ Where ∆Go is the standard free energy change of the reaction. At equilibrium ∆G = 0 .
⚫ ჻ ∆Go = - RT ln {[(PL)l (PM)m]/[(PA)a (PB)b]}
⚫ Since ∆Go for a particular reaction at a given temperature is a fixed quantity, the logarthimic factor must
have constant value. It is customary to call this conatant as “equilibirium constant”, KP.
⚫ i.e {[(PL)l (PM)m]/[(PA)a (PB)b]} = KP
⚫ Now,
⚫ ∆Go = -RT ln KP
⚫ or ∆Go = -2.303 RT log KP

37. Van’t Hoff Isochore (Van’t Hoff equation)
⚫ The variation of equilibirium constant with temperature is given by Van’t
Hoff Isochore. It is also known as “Van’t Hoff equation”.
⚫ The Van’t Hoff reaction isotherm is
⚫ ∆Go = -RT ln KP...........(1)
⚫ Differentiating w.r. to temperature at constant pressure
⚫ ə(∆Go)/əTP = - R ln KP – RT [ə(ln KP)/əT]P
⚫ Multiplying throught by T
T[ə(∆Go)/əTP]= - RT ln KP – RT2 [ə(ln KP)/əT]P........(2)
⚫ Comparing eqn (1) and (2)
⚫ T[ə(∆Go)/əTP]= ∆Go– RT2 [ə(ln KP)/əT]P.............(3)
⚫ From Gibbs-Hhelmholtz equation
⚫ ∆Go = ∆Ho + T [ə(∆Go)/ƏT]P
⚫ or T [ə(∆Go)/ƏT]P = ∆Go - ∆Ho
⚫ Substituting the value of T [ə(∆Go)/ƏT]P in eqn (3)
⚫ ∆Go - ∆Ho = ∆Go– RT2 [ə(ln KP)/əT]P
⚫ or RT2 [ə(ln KP)/əT]P = ∆Ho
⚫ or [ə(ln KP)/əT]P = ∆Ho/RT2
⚫ or [d(ln KP)/dT]P = ∆Ho/RT2
⚫ This is the differential form of Van’t Hoff isochore.

38. Integrated form of Van’t Hoff Isochore
⚫ [d(ln KP)/dT]P = ∆Ho/RT2
⚫or d(ln KP) = (∆Ho/R).(dT/T2)
⚫Integrating between limits
⚫ 1ʃ2d(ln KP) = (∆Ho/R)T1ʃT2(dT/T2)
⚫ Ln (KP)2/ (KP)1 = -(∆Ho/R) [-1 /T] T1
T2
⚫ = - ∆Ho/R[(1/ T2)- [(1/T1)]
⚫ = ∆Ho/R[(1/ T1)- [(1/T2)]
⚫ = ∆Ho/R[(T2 - T1)/( T1T2)]
⚫or2.303 log (KP)2/ (KP)1 = ∆Ho/R[(T2 - T1)/( T1T2)]
⚫჻ log (KP)2/ (KP)1 = (∆Ho/2.303R)[(T2 - T1)/( T1T2)]

39. Applications
1. The Van’t Hoff reaction isochore permits the calculation of the
enthalpy of a reaction (∆H) when the equilibrium constants at
two different temperatures are known. It can also be evaluated
by plotting log KP against 1/T. A straight line is obtained. The
slope of the straight line is equal to - ∆Ho/2.303R.
2. If the equilibrium constant of a reaction at one temperature of
known. That of another temperature can be calculated.
3. It helps to calculate the heat of solution from the solubilities of a
substance at two different temperatures.
log S2/S1 = (∆H/2.303R)[(T2 - T1)/( T1T2)]
4. It is useful to find out the heat of dissociation of weak
electrolytes from the temperature co-efficient of the dissociation
constant.
5. The Van’t Hoff isochore may be regarded as a quantitative
expression of Le Chatlier’s principle.

40. Maxwell’s relationships
⚫ The state functions E, H, S, A and G are related to P, V and T by means of the following
fundamental equations:
⚫ i) dE = TdS- PdV
⚫ ii) dH = TdS+ VdP
⚫ iii) dA = -PdV- SdT
⚫ iv) dG = VdP- SdT
⚫ From the fundamental equations, the following four relations which are extensively used in
thermodynamic calculations can be derived. These are called “Maxwells equations or
relationships”.
⚫ Consider the eqn.
⚫ dE = TdS - PdV
⚫ If V is constant, dV = 0
⚫ ჻ dE = TdS
⚫ or (əE/əS)V = T ..................(1)
⚫ If S is constant, dS = 0
⚫ ჻ dE = - PdV .................(2)
⚫ or (əE/əS)S = -P
⚫ Differentiating eqn. (1) w.r.to V at constant S and eqn.(2) w.r.to S at constant V, we get
⚫ (ə2E/əSəV) = (əT/əV)S ..................(3)
⚫ (ə2E/əVəS) = - (əP/əS)V ..................(4)
⚫ Comparing eqns (3) and (4)
⚫ (əT/əV)S = - (əP/əS)V ..................(5)
⚫ Following the same procedure, we have
⚫ (əT/əP)S = (əV/əS)P from(ii) ..................(6)
⚫ (əS/əV)T = (əP/əT)Vfrom(ii) ..................(7)
⚫ (əS/əP)T = - (əV/əT)Pfrom(ii) ..................(8)

41. The Third Law of Thermodynamics
⚫Nernst Heat Theorem
⚫ The Gibbs-Helmholtz equation may be used to calculate ∆H by
measurment of ∆G at different temperatures.
⚫ ∆G = ∆H + T [ə(∆G)/ƏT]P
⚫ Since ∆H can also be dteremined readily from heat capacity
measurments, the calculation of ∆G from such thermal data was
attempted by T.W.Richards, E.Nernst and others. Through a series of
experiments on galvanic cells, Richards found that the value of [ə(∆G)/ƏT]P
decreases as the temperature is lowered consequently ∆G and ∆H approach
each other and becomes equal at the absolute zero of temperature.

42. This observation was generalised in the form of
“Nernst heat theorem”. According to the theorem
“[ə(∆G)/ƏT]P decreases with temperature asymptotically and
equals to zero in the vicinity of absolute zero temperature”.
This means that ∆G and ∆H will have the same value near 0o K.
Mathematically, the heat theorem is written as
Lt [ə(∆G)/ƏT]P = Lt [ə(∆H)/ƏT]P = 0
T→0 T→0
From Second law of thermodynamics,
[ə(∆G)/ƏT]P = - ∆S
[ə(∆H)/ƏT]P = ∆CP
Thus, it follows from the heat theorem that
Lt ∆S = 0
T→0
Lt ∆CP = 0
T→0

43. Applications of Nernst Heat Theorem
⚫ The heat theorem enables one to determine the free energy change from
thermal data.
⚫ ∆CP = [ə(∆H)/ƏT]P ...........Kirchoff’s equation
⚫ But ∆H = T ∆S
⚫ ჻ ∆CP = T[ə(∆S)/ƏT]P
⚫ or [ə(∆S)/ƏT]P = (∆CP)/T
⚫ or d(∆S) = (∆CP) . (dT/T)
⚫ = (∆CP) (dlnT)
⚫ General integration of the above eqn. Gives
⚫ ∆S = ∆CP lnT + C
⚫ where C is the integration constant. When T = 0, ∆S = 0[Nernst heat theorem].
⚫ Now,
⚫ ∆S = ∆CP .lnT
⚫ Since ∆G = ∆H - T∆S
⚫ ∆G = ∆H - T (∆CP .lnT)
⚫ Limitations of Nernst heat theorem
⚫ Nernst assumed that the heat theorem is applicable to all condensed systems
including solutions. But Max Planck pointed out that solutions at 0o K have a
positive entropy equal to entropy of mixing. Experiments proved that the heat
theorem is applicable to “perfect crystalline solids” only.

44. Third law of Thermodynamics
⚫ Max Planck in 1912 extended Nernst heat theorem and enunciated
the third law of thermodynamics as “the entropy of a pure solid or
liquid becomes zero at the absolute zero of temperature”.
Mathematically
⚫ Lt S = 0
T→0
Lewis and Gibson pointed out that supercooled liquids such as
glasses retain positive entropies as the temperature approaches
absolute zero. For this reason, Lewis and Randall modified the third
law as
“Every substance has a finite positive entropy, but at the absolute
zero of temperature the entropy may become zero and it does become
zero in the case of a perfectly crystalline substance”.
Applications of the Third law of Thermodynamics
i) It enables us to calculate the absolute entropy of a solid, liquid or gas.
ii) It is used to evaluate the entropy change in chemical reactions.
iii) It is used to calculate the free energy change in chemical reactions
from thermal data.

45. Determination of Absolute entropy of Solids
⚫ The entropy change is given by
⚫ dS = δq/T
⚫ At constant pressure, δq = dH
⚫ ჻ dS = dH/T
⚫ or (əS/əT)P = (əH/əT)P . 1/T
⚫ By definition, (əH/əT)P = CP
⚫ ჻ (əS/əT)P = CP/T
⚫ Or dS = CP . dT/T
⚫ On integration between the limits
⚫ 0ʃTdS = 0ʃTCP dlnT
⚫ when T=0, the absolute entropy S0 = 0 for a pure crystalline solid
⚫ ჻ ST = 0ʃTCP dlnT
⚫ = 0ʃT2.303CP d logT
⚫ where ST is the absolute entropy of the solid at a given temperature T. It can be
evaluated graphically as shown below :
⚫ The values of CP are measured at various temperatures as low as 15o K and the
plotted against log T. The curve is extrapolated to 0o K and the area under the
curve gives the value of the integral which when multiplied by 2.303 gives the
absolute entropy ST.